Question

Solve the following inequality:
${\log }_{0.5}(x+5{)}^2>{\log }_{1/2}(3x-1{)}^2$

Answer 1

${\log }_{0.5}{(x+5)}^2>{\log }_{1/2}{(3x-1)}^2$

${(x+5)}^2<{(3x-1)}^2$ also, $x+5>03x-1>0$

${(x+5)}^2-{(3x-1)}^2<0x>-5x>\frac{1}{3}$

$[x+5-(3x-1\left)\right][x+5+3x-1]<0$

$(x+5-3x+1)(4x+4)<0xϵ(-5,\infty )$

$xϵ(\frac{1}{3},\infty )$

$(-2x+6)(4x+4)<0$

$(x-3)(x+1)>0$

$xϵ(-\infty ,-1)\bigcup (3,\infty )$

Taking intersection we get $xϵ(3,\infty )$